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Proof step 3 from proof rmpos-in (previous step)

n < # x ⊦ a ∈ x -p n ↔ a ≠ x[n] ∧ a ∈ x, x = [], dups(x)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
n < #(a0 + x0), ∀ n, a. n < # x0 → (a ∈ x0 -p n ↔ a ≠ x0[n] ∧ a ∈ x0) ∨ x0 = [] ∨ dups(x0) ⊦ a0 + x0 = [], a ∈ (a0 + x0) -p n ↔ a ≠ (a0 + x0)[n] ∧ a ∈ a0 + x0, dups(a0 + x0)

Rule argument

term: x

Additional information

Used simplifier rules: list freely generated by [], +;;